Problem: Divide the following complex numbers. $\dfrac{8-2i}{-4+i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-4-i}$. $ \dfrac{8-2i}{-4+i} = \dfrac{8-2i}{-4+i} \cdot \dfrac{{-4-i}}{{-4-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(8-2i) \cdot (-4-i)} {(-4)^2 - (i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(8-2i) \cdot (-4-i)} {(-4)^2 - (i)^2} $ $ = \dfrac{(8-2i) \cdot (-4-i)} {16 + 1} $ $ = \dfrac{(8-2i) \cdot (-4-i)} {17} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({8-2i}) \cdot ({-4-i})} {17} $ $ = \dfrac{{8} \cdot {(-4)} + {-2} \cdot {(-4) i} + {8} \cdot {-1 i} + {-2} \cdot {-1 i^2}} {17} $ $ = \dfrac{-32 + 8i - 8i + 2 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{-32 + 8i - 8i - 2} {17} = \dfrac{-34 + 0i} {17} = -2 $